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Energetics of metabolism

Dr. Bódis Emőke

October 7, 2015

JJ9

Why do we study difficult thermodynamics?

The laws and principles of the thermodynamics describe the characteristics of matter- and energy flow of the different biochemical systems.

- From these we can determine the direction of a spontaneous reaction.

The 0. law of the thermodynamics (equilibrium and stability)

If system A is in thermal equilibrium with system C, and system B is in thermal equilibrium with system C, then system A is in thermal equilibrium with system B.

Two systems are said to be in thermal equilibrium if, when they are placed in ``thermal contact'' (basically, contact that permits the exchange of energy between them), their state variables do not change.

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The I. law of the thermodynamic (conservation of energy)

ΔU = U2-U1 = Q + W

Energy can be converted from one form to another with the interaction of heat, work and internal energy. Energy can be changed from one form to another, but it cannot be created or destroyed.

BUT! In biological systems instead of internal energy it is more useful to apply ENTHALPY.

H = U + pVΔH = ΔU + pΔV = Q + W + pΔV = Q - pΔV + pΔV = Q

ΔH: Enthalpy change is the amount of heat content used or released in a system at constant pressure. (ΔU: Internal energy change is the amount of heat content used or released in a system at constant volume. )

HpVU

The II. Law of the thermodynamics (The direction of the processes)

Systems tend to proceed from ordered (low-entropy or low-probability) states to disordered (high-entropy or high-probability) states.

ΔQ = T ΔS

Heat transferred in a process change the entropy of the system.

Entropy represents energy dispersion:large number of molecular motions is relatable to quantized states (microstate)

S = k ln W = k ln (W2-W1)

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A termodinamika III. főtétele

Tökeletes kristalyos, tiszta anyagok entrópiaja (vegyuleteke is!) T = 0 K-en nulla.

Következmenye: T= 0 K-en ismerjuk az entrópia abszolut erteketDE! Biológiai folyamatokban az entrópia valtozas hasznosabb, mint az abszolut entrópia.

Pl: ΔG = ΔH – TΔS

A III. főtetel statisztikus magyarazata: S = k ln W egyenletből T = 0 K hőmersekleten a tökeletesen kristalyos anyagok reszecskeimind a leheto legalacsonyabb energiaju allapotban vannak

W = 1 es S = 0

A legalacsonyabb mert hőmerseklet a termeszetben 1 K (Boomerang csillagködben, 5000 fenyevnyire tőlunk, a Kentaur csillagkepben)

Egy reakció csak akkorspontan, ha a szabad energia

negatív!

It is usefull to apply for biological systems.

Question: Will the reaction proceed in the direction written?

1. The thermodynamic condition: Gibbs free energy must decrease.

Gibbs free energy: measures the "usefulness" or process-initiating work obtainable from a thermodynamic system at constant temperature and pressure.

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2. The reaction kinetic condition: the reactants must decrease the activation energy

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The catabolic ways are convergent, the anabolic ways are divergent.

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First period of the cellular respiration: Glycolysis

aroses 2 pyruvate molecules.

From 1 glucose molecule

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Pyruvate, a versatile metabolite

The final product of glycolysisCan be used in several ways

Oxygen is present (aerobic condition): pyruvate is oxidized and will be the acetyl group of acetyl-CoA and will further metabolized in the tricarboxylic acid cycle

Oxygen is absent (anaerobic condition): - lactic acid fermentation- alcoholic fermentation

Glikolízis, tejsavas fermentació

C6H12O6 2 H3C - CHOH – COO- +2 H+ ΔG = - 183,6 kJ/mol

Emellett netto 2 ATP is keletkezik

2 ADP + 2 Pi 2 ATP + 2 H2O ΔG = 61 kJ/mol

ΔG = - 183,6 + 61 = - 122,6 kJ/mol

A szabad energia valtozas bőven fedezi a 2 ATP szinteziset:

(61 / 183,6) * 100% = 33,2 %

(A felszabaduló energia 33,2 %-a fordítódik a 2 ATP szintezisere.)

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Why does Glycolysis occur? Are the thermodynamic and kinetic conditions necessary for the reactions met?

The first phase of Glycolysis

The first phase of Glycolysis: First energy priming reaction

Kinase: enzyme responsible for transferring the γ-phosphate of ATP to moleculesHexokinase: phosphorylate hexose sugars (C-6)

The first and third steps of the Glycolysis pathway will be primed with energy from hydrolysis of 2 ATP so that at the end of Glycolysis 4 ATP will be formed (net 2 ATP)

ATP cleavage: ΔG = -30,5 kJ/molPhosphorylation: ΔG = 13,8 kJ/molFirs step: ΔG = -30,5 + 13,8 = - 16,7 kJ/mol

Water pump has to be primed with small amount of water to deliver more water out.

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Phosphorylation of glucose to glucose-6-phosphate

Hexokinase (enzyme)

GlucoseMg2+ - ATP

- phosphorylation: glucose glucose-6-phosphate

- Conformational change in the presence of Mg2+ -ATP and glucose (open-closed)

- Regulation: The enzyme function is inhibited allosterically in the presence of high concentration of glucose-6-phosphate

glucose-6-phosphate is a the brancs point of several metabolic pathways.

Glucokinase: hexokinase in liver and pancreas

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The first phase of Glycolysis: glucose to glucose-6-phosphate

The reaction occur, because

1. The thermodynamic condition meets (ΔG = - 16,7 kJ/mol)

2. The kinetic condition meets, because the presence of hexokinase decrease the activation energy of the step.

Gibbs free energy change during the whole process of Glycolysis

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Third step of Glycolysis

Third step of Glycolysis: Second energy priming reaction

Catalysator: Phosphofructokinase ATP hydrolysis: ΔG = -30,5 kJ/mol2. Phosphorylation: ΔG = 11,7 kJ/molThird step: ΔG = -30,5 + 13,8 = - 18,8 kJ/mol

Regulation of the protein:

- 2 ATP binding sites- Low concentration of ATP: ATP binds to the high

affinity binding site (ATP at active site)- High concentration of ATP: : ATP binds also to the

low affinity binding site (ATP at regulatory site) glycolysis “turns off”

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Fourth step of Glycolysis

Fourth step of Glycolysis: concentration-dependent reaction

Fructose-Bisphosphate aldolase: claves between C3-C4 carbons yield two triose phosphates

Concentration-dependent reaction: - in vitro: the reaction does not proceed effectively from left to right

(ΔG positive: 23,97 kJ/mol)- in vivo: efficient

(ΔG negative: -0,23 kJ/mol

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Seventh step of Glycolysis

ADPMg2+

phosphate

Seventh step of Glycolysis: Synthesis of the first two ATPs

- Phosphoglycerate kinase: phophorylation of ADP (substrate-level phophorylation )- Two ATP were consumed in the first phase of Glycolysis , the Phosphoglycerate kinase reaction “pays off” the ATP debt created by the priming reactions- In living cells: ΔG : -0,1 kJ/mol

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Tenth (last) step of Glycolysis

Tenth (last) step of Glycolysis: Synthesis of the second two ATPs

Catalysator : Pyruvate kinasePresence of Mg2+ es K+ are necessary 1. step: phosphorylation of ADP2. step: enol-ketone conversion

Highly favorable and spontaneous conversion of the enol to the more stable keto form following the phosphorylation group transfer step.

ΔG = -31,7 kJ/mol

Gibbs free energy “pays” the production of ATP.

ΔG = 30,5 kJ/mol

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Thank you!